# Math Help - Minimizing Length of a Crease

1. ## Minimizing Length of a Crease

I have been working on this problem for a while now, i have nailed the other 2 cases possible for this problem but the last case has me stuck

here is the problem

You have a sheet of paper that is 6 units wide and 25 units long placed
so that the short side is facing you. Fold the lower right hand corner
over to touch the left side. Your task is to fold the paper in such a way
that the length of the crease is minimized. What is the length of the
crease?

Do with without using trigonometry

currently i am working on the case on the far right (i have completed the first two)

This is the diagram that i have made.

Everything that is not in computer font was what i originally had drawn myself. The computer additions are what my teacher added to my diagram, all of which have confused me even more.

My professor has told me that i have to relate L (length of the crease) in terms of x. He has also given me A which he says i will needed. I dont know how these variables relate or how to start this problem

If any more diagrams are needed i will provide them on request

This is urgent i need to turn this in asap

Thanks for taking your time to look and help me

2. Originally Posted by kaistylez
I have been working on this problem for a while now, i have nailed the other 2 cases possible for this problem but the last case has me stuck

here is the problem

[B] You have a sheet of paper that is 6 units wide and 25 units long placed
so that the short side is facing you. Fold the lower right hand corner
over to touch the left side. Your task is to fold the paper in such a way
that the length of the crease is minimized. What is the length of the
crease?

Do with without using trigonometry
...
If any more diagrams are needed i will provide them on request
Hello,

I've modified your diagram a little bit (see attachment)

1. If you put the right bottom corner on the left top corner then there is nothing to minimize.

2. You can prove that the 2 red right triangles are congruent.

3. Thus all lines coloured blue have the same length. In your diagram one of the blue lines was labeled y.

4. You get 2 equations:
[1] $y = 25 - t$

[2] $y^2 = 6^2 + t^2$ Plug the term for y from [1] into [2]:

$(25-t)^2=36+t^2\ \Longleftrightarrow \ t = \frac{589}{50} = 11.78$

3. Hello,

it's me again.

As I've shown t = 11.78.

You can prove that a = 25 - 2t = 1.44

From your diagram you know that

$L^2 = a^2 + 6^2\ \Longrightarrow \ L = \sqrt{1.44^2 + 36} \approx 6.17$

I've attached a (nearly) exact drawing of the piece of paper folded as it is stated in the problem.

4. Thank you for the reply

i showed this to my professor and he had this to say.

"from what I see you got a specific answer and that shouldn't happen unless you're assuming the case with the lower right corner touching the upper left corner. I think you may be assuming this in one of your equations involving t. You still need to generate the general formula relating the crease length to the position of the corner along the left hand margin - and perform some analysis (calculus or otherwise) on it."

SO would i achieve the generallization if i changed y=25-t to y=x-t?

5. Originally Posted by kaistylez

i showed this to my professor and he had this to say.

"from what I see you got a specific answer and that shouldn't happen unless you're assuming the case with the lower right corner touching the upper left corner. I think you may be assuming this in one of your equations involving t. You still need to generate the general formula relating the crease length to the position of the corner along the left hand margin - and perform some analysis (calculus or otherwise) on it."

SO would i achieve the generallization if i changed y=25-t to y=x-t?
Hello, kaistylez,

1. The result which I send you in my previous posts is already the minimum crease.

2. It is not necessary to get any generalizations because you can show that by some plausible considerations:

If you fold the paper the crease is the angle bisector of two legs which have the same length. They were labeled y and I have marked it blue in my previous posts.
If the angle between these 2 y's is exactly 90° the the crease has a maximum and a length of $L=\sqrt{6^2 +6^2}=6 \cdot \sqrt{2} \approx 8.485$

I labeled the starting point of the crease with C. If C is going up on the right side of the paperstrip then the angle between the two y's is increasing and the crease is decreasing. You'll get the shortest possible distance for the crease if the bottom right corner is placed over the top left corner.

I've documented 4 different stages of the position of C. The blue line is the crease which is obviously decreasing in length.

6. thank you again for the input onto this problem.

After showing this to my professor. He says that this is good proof, but i still need to prove my assertion by math (calculus). With this being said the only way i can think about it is to take the derivative of the L^2 = {[(36-x^2)/2]+25}^2 +6^2. i found this a length by replacing 25 with x in this equation $y = 25 - t$ i get $y = x - t$ when i solve for t i get t = [(36-x^2)/-2x] then i went onto subbing that into the equation a = 25 - 2t, i then get a = 25 + [(36-x^2)/x]. Finally i sub it into the L^2 equation. But when i take the derivative i get something really long and ugly.

My professor tells me that by proving my assertion with calculus i can finally be done with this problem.

7. Originally Posted by kaistylez
thank you again for the input onto this problem.

After showing this to my professor. He says that this is good proof, but i still need to prove my assertion by math (calculus).
...
My professor tells me that by proving my assertion with calculus i can finally be done with this problem.
Hello, kaistylez,

it's me again - and I hope that I'm not too late.

I've attached a drawing of the folded paper.

I labeled
the green line t
the blue line y
and the length of the crease L in red.

If you fold the bottom right point to the left margin then this point is x over the bottom.

1. Domain: $6 \leq x \leq 25$

2.
$x = y + t \Longleftrightarrow t = x - y$ [1]
$y^2 = 6^2 + t^2 \Longrightarrow y^2 = 36 + (x-y)^2 \Longrightarrow y=\frac{36+x^2}{2x}$ [2]

$L^2=6^2 + (y-t)^2 \Longrightarrow L^2=36+(2y-x)^2$ using equation [1]

$L^2=36+\left(2\cdot \frac{36+x^2}{2x}-x \right)^2 = 36 + \left(\frac{36}{x} \right)^2 = 36 + \frac{1296}{x^2}$ Therefore your function is:

$L(x) = \sqrt{36 + \frac{1296}{x^2}}$ (By the way: You can check this equation: Plug in x = 25 then you should get the result which you already know)

3. Calculate the first derivation: After a few steps you should get:

$L'(x) =-\frac{216}{x^2 \cdot \sqrt{x^2+36}}$

This fraction is never zero for any real x.
Therefore the extrema are at the bounds of the domain: The smaller x is the greater is L - and: the greater x is the smaller is L.

You get the maximum length if x = 6 then $L = 6 \cdot \sqrt{2} \approx 8.485$ and you get the minimum length if x = 25 then $L=\sqrt{\frac{23796}{625}} \approx 6.17$

8. Hello, kaistylez,

I noticed that I havn't attached the drawing on which my calculations are based. So now here it comes:

9. ## Re: Minimizing Length of a Crease

Only in case it isn't quite clear which geometric properties I used I've attached a more detailed sketch.

Prove that the angles marked in orange are equal. Show that the grey triangle is isesceles because the base angles at the red base are equal.